TITLE: Does a surjective linear map commute with the interior of a convex body?
QUESTION [6 upvotes]: Let $T:\Bbb R^n \to \Bbb R^m $ be a surjective linear map.
Suppose $A\subset\Bbb R^n$ is convex, compact, $0\in \operatorname{int}(A)$, and centrally symmetric.
Is it true that $T(\operatorname{int}(A))=\operatorname{int}(T(A))$?.
Notes: 

$\operatorname{int}(S)$ means interior of $S$ in the topology of $\mathbb{R}^n$ or $\mathbb{R}^m$.
Convexity is important: e.g., consider the image of sphere $S^2\subset \mathbb{R}^3$ under projection
$A$ having nonempty interior is also important: consider the image of a disk $D^2\subset \mathbb{R}^3$ under the same projection.

REPLY [3 votes]: Since $T$ is surjective, it is open. Therefore $T(\operatorname{int} A)$ is an open subset of $T(A)$, so $T(\operatorname{int} A) \subseteq \operatorname{int} T(A)$.
Pick an $a_0 \in \operatorname{int} A$, and choose an arbitrary $y \in \bigl(\operatorname{int} T(A)\bigr) \setminus \{T(a_0)\}$. Then there is an $\varepsilon > 0$ such that $z = T(a_0) + (1+\varepsilon)\bigl(y - T(a_0)\bigr) \in \operatorname{int} T(A)$. Choose $x \in A \cap T^{-1}(\{z\})$. Then the "tipless cone"
$$C = \{ tx + (1-t)a : t \in [0,1),\, a \in \operatorname{int} A\}$$
is open, and by convexity contained in $A$, hence $C \subseteq \operatorname{int} A$. We have
$$w = a_0 + \frac{1}{1+\varepsilon}(x - a_0) \in C$$
and
\begin{align}
T(w) &= T(a_0) + \frac{1}{1+\varepsilon}\bigl(T(x) - T(a_0)\bigr) \\
&= T(a_0) + \frac{1}{1+\varepsilon}\bigl(z - T(a_0)\bigr) \\
&= T(a_0) + \frac{1}{1+\varepsilon}(1+\varepsilon)\bigl(y - T(a_0)\bigr) \\
&= y,
\end{align}
showing $y \in T(\operatorname{int} A)$. Since clearly also $T(a_0) \in T(\operatorname{int} A)$, it follows that $\operatorname{int} T(A) \subseteq T(\operatorname{int} A)$.
We note that the proof only used that $T$ is an open linear map, and $A$ convex with nonempty interior, so the result also holds in an infinite-dimensional setting.